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Vikentia [17]
3 years ago
9

CAN SOMEONE PLEASE HELP ME???

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

Option D, x = 4

Step-by-step explanation:

Option A:  y = 4 doesn't work because that line would be horizontal

Option B:  y = 4x doesn't work because that would be diagnol

Option C:  x = -4 doesn't work because that would a vertical line at -4

<em>Option D:  x = 4 works because that would a vertical line at 4</em>

<em />

Answer:  Option D, x = 4

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Answer:

x=-3

Step-by-step explanation:

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A university found that 10% of its students withdraw without completing the introductory statistics course. assume that 20 stude
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Two coins, A and B, each have a side for heads and a side for tails. When coin A is tossed, the probability it will land tails-s
BlackZzzverrR [31]

Answer:

Is the number of tosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?

b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.

Step-by-step explanation:

a) Data and Calculations:

The probability of coin A landing tails-side up = 0.5

The proportion of times coin A lands tails-side up (PA) = 20 * 0.5 = 10

Therefore, the probability of coin A landing heads-side up = 0.5 (1 - 0.5)  

And the proportion of times that coin A lands heads-side up = 20 * 0.5 = 10.  

The proportion on either side is equally distributed.

This is why 20 tosses for coin A is enough, since the sample proportions PA is approximately normal, symmetric, and equally distributed.  There will be equal amounts of 10 tosses (0.5 *20) for either heads-side up or tails-side up.

For coin B, the probability of landing tails-side up = 0.8

The proportion of times coin B lands tails-side up (PB) = 20 * 0.8 = 16

Therefore, the probability of coin B landing heads-side up = 0.2 (1 - 0-.8)

The proportion on either side is not equally distributed, but skewed.

This is why 20 tosses for coin B is not enough, since the sample proportions PB is not approximately normal, symmetric, and equally distributed.  There will be 16 tosses landing tails-side up (0.8*20) and only 4 tosses landing heads-side up (0.2*20).

6 0
3 years ago
Solve the inequality <br> 5x+2 &lt; 32
AnnyKZ [126]

Answer:

x<6

Step-by-step explanation:

First, let's work on the left hand side of your inequality, the 5x+2

This means, for instance, to see if it can be simplified at all.

Multiply x and 5

Multiply x and 1

The x just gets copied along.

The answer is x

x

5*x evaluates to 5x

5*x+2 evaluates to 5x+2

So, all-in-all, the left hand side of your inequality can be written as: 5x+2

Now, let's work on the right hand side of your inequality, the 32

The right hand side of your inequality can be written as: 32

So with these (any) simplifications, the inequality we'll set out to solve is:

5x+2 ‹ 32

Move the 2 to the right hand side by subtracting 2 from both sides, like this:

From the left hand side:

2 - 2 = 0

The answer is 5x

From the right hand side:

32 - 2 = 30

The answer is 30

Now, the inequality reads:

5x ‹ 30

To isolate the x, we have to divide both sides of the inequality by the other "stuff" (variables or coefficients)

around the x on the left side of the inequality.

The last step is to divide both sides of the inequality by 5 like this:

To divide x by 1

The x just gets copied along in the numerator.

The answer is x

5x ÷ 5 = x

30 ÷ 5 = 6

The solution to your inequality is:

x ‹ 6

8 0
2 years ago
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